Chọn D
Ta có f'\left(x\right)=12x^{2} -3
\(f'\left(x\right)=0\Leftrightarrow \left[\begin{array}{l} {x=\frac{1}{2} \in \left[\frac{1}{4} ;\frac{4}{5} \right]} \\ {x=-\frac{1}{2} \notin \left[\frac{1}{4} ;\frac{4}{5} \right]} \end{array}\right. .\)
\(f\left(\frac{1}{4} \right)=-\frac{27}{16} , f\left(\frac{1}{2} \right)=-2, f\left(\frac{4}{5} \right)=-\frac{169}{125} .\)
Do đó \({\mathop{{\rm max}}\limits_{\left[\frac{1}{4} ;\frac{4}{5} \right]}} f\left(x\right)=-\frac{169}{125} =M, {\mathop{{\rm min}}\limits_{\left[\frac{1}{4} ;\frac{4}{5} \right]}} f\left(x\right)=-2=m\)
Vậy \(M+m=-\frac{419}{125} \)